I am looking for a set of differential equations (to be solved numerically for an educational program) that would describe the position and apparent time of an accelerated clock relative to a non-inertial reference frame. For now, all movement is along a single spatial dimension. I am not interested in considering the time light takes to travel between observers; assume that they are clever enough to account for this delay. As inputs to this system of equations, I would have:

  • An arbitrary, time-dependent acceleration a(t) as experienced by an observer in the non-inertial frame. In practice, this would be a simple list such as a(t=0) = 0; a(t=1) = 0.2; ...
  • The initial position x(t=0), clock time t'(t=0), and velocity v(t=0) of the accelerated clock as seen in the non-inertial reference frame
  • Another time-dependent acceleration a'(t') which describes the acceleration experienced by an observer traveling with the clock.

To make this (hopefully) more clear, I'll give an example: an observer A, initially at rest, sees a relativistic rocket pass by carrying observer B. When A's watch reads t=0, she notes that B's position is x(t=0), B's speed is v(t=0) and B's (prominently displayed) clock reads t'(t=0). A has a list of instructions a(t) which tell her to set her rocket engines to achieve a specific acceleration at specific times. B has a similar set of instructions a'(t'). Both A and B consult their own watches when determining when to change their acceleration.

I have seen this question: What is the displacement of an accelerated and relativistic object?, which tells me enough to model the situation for an inertial observer, but I have not figured out how to reapply this for a non-inertial observer. Similarly, I can compute the worldlines of non-accelerated objects from the perspective of a non-inertial observer, but it seems something sneaky happens when both the observer and observed are accelerating.

I have also tried computing both worldlines from an inertial reference frame and then using the Lorentz transformation [x' = g (x - vt); y' = g (t - vx/c^2)] to ask what B's worldline would look like from A's perspective, but this did not work (I can describe more about this, but for now it should suffice to say that the worldlines were not remotely correct).


I would approach this by specifying a third observer C who is in an inertial frame. It's relatively easy to calculate the time of an accelerated frame wrt the inertial frame, and vice verse. So if you take some proper time for A, you can convert this into the time for the inertial observer then convert it again into the proper time for observer B.

John Baez's article on the relativistic rocket gives the equations for converting from the accelerated time to observer C's time and vice versa. However note that these are the equations for constant acceleration. If you want to model time dependant accelerations then You'll need to delve into chapter 6 of Gravitation by Misner, Thorne and Wheeler.

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  • $\begingroup$ Thanks, I'll look through Gravitation and see where that leads me. $\endgroup$ – Luke Aug 16 '12 at 15:14
  • $\begingroup$ Good luck, it's not an easy read! $\endgroup$ – John Rennie Aug 16 '12 at 15:53

There is no such thing as a non-inertial reference frame in SR at all, actually. Attempts to build such a frame (Rindler coordinates) meet some essentially GRstic issues, such as imcompleteness, event horizon, particle horizon, coordinate singularity, some peculiar (gravitationally-affected) kinematics and physics as seen by the observer.

However, non-inertial bodies and non-inertial observers can be described with no problem. For the observer, the picture s/he sees around herself/himself could be most easily calculated, if you assume the "instantly comoving" inertial frame, that is the frame moving at the same velocity as the observer at taken instant. (See also here.) This is exactly adequate, since the observer could switch off her/his jet for a tiny instant, and become inertial. Though you should be careful in which values to consider in such a frame. Some things are immediately observable, such as photons arrived from th space, and the apparent position and velocity of the other rocket as seen by these photons. They are physical and could be measured with instruments. And also there are unobservable things, like the "simultaneous position and velocity" of the distant rocket, which could not be seen and could only be calculated beforehand, based on the information the observer has an access to. If the information (for example, about future accelerations of the distant rocket) is unavailable, then such things could be calculated by no means.

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  • $\begingroup$ "There is no such thing as a non-inertial reference frame in SR at all" This is wrong. SR doesn't have any problem with non-inertial reference frames. You do get effects such as horizons, but this is not a problem for SR, nor does it imply the need for GR. GR is only needed if there is spacetime curvature. Einstein did originally think of GR as a generalization of SR to noninertial frames, but it turns out that that was simply a mistake. What do you mean by "incompleteness?" Geodesic incompleteness? If so, then that's incorrect. $\endgroup$ – user4552 Aug 12 '13 at 14:16
  • $\begingroup$ I mean all the issues GR meets when one tries to cover the spacetime manifold with coordinate maps. These issues come from the fact that manifold-to-map is not one-to-one correspondence, and they are not easier in the case of a globally trivial manifold. The manifold is not covered by one map, Jacobian can become zero, and even one point can be given several coordinates. Light cone can get unusual arrangement, with non-timelike 0th axis, and/or non-spacelike 1,2,3rd axes. All these effects, though not dealing specifically with gravitational physics, are covered by GR textbooks, and not SR ones $\endgroup$ – firtree Aug 22 '13 at 5:31
  • $\begingroup$ GR is not as narrow as the Einstein's equation and the curved spacetime case. It also sets some conceptual framework which previous theories (including SR) do not need. And it is this framework that is needed for this question, not the Einstein's equation. $\endgroup$ – firtree Aug 22 '13 at 5:32

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